Superposition of Magnetic Fields

[latentcorpse]

this problem relates to finding the on axis field of two coaxial coils of the same radius when they have their currents in the same/opposite directions.
the field from one coil is

B_z=\frac{\mu_0}{2} I R^2 \frac{R^2}{(z^2+R^2)^{3/2}}

now the field from a coil looks like that of a dipole i.e. it's symmetric above and below the axis. so if the coils have radius R and separation 2d and i want the field a smaoll distance m from the midpoint and the currents are in the same direction i can get the field from the top as

\frac{\mu_0}{2} I R^2 \frac{R^2}{((z-m)^2+R^2)^{3/2}}
and from the bottom as
\frac{\mu_0}{2} I R^2 \frac{R^2}{((z+m)^2+R^2)^{3/2}}

and then hopefully superimpose.
but this is where I am having difficulty making it into all one term (i don't want to leave my final answer as something + something else)
since m is small am i meant to taylor expand or something?

 

[Jhero]

Hello there,

Thank you for bringing up this interesting problem. I can see where you are having difficulty in combining the two terms into one. In this case, you can use the principle of superposition to combine the two fields into one. Since the fields from the two coils are symmetrical, we can take the average of the two fields to get the total field at a distance m from the midpoint.

So, the total field at a distance m from the midpoint can be written as:

B_z = \frac{1}{2} \left[ \frac{\mu_0}{2} I R^2 \frac{R^2}{((z-m)^2+R^2)^{3/2}} + \frac{\mu_0}{2} I R^2 \frac{R^2}{((z+m)^2+R^2)^{3/2}} \right]

Simplifying this expression, we get:

B_z = \frac{\mu_0}{4} I R^2 \left[ \frac{R^2}{((z-m)^2+R^2)^{3/2}} + \frac{R^2}{((z+m)^2+R^2)^{3/2}} \right]

Now, we can use the binomial expansion to simplify the terms in brackets. Expanding up to the second order, we get:

B_z = \frac{\mu_0}{4} I R^2 \left[ \frac{1}{R^3} \left( \frac{1}{(1+\frac{(z-m)^2}{R^2})^{3/2}} + \frac{1}{(1+\frac{(z+m)^2}{R^2})^{3/2}} \right) \right]

Using the binomial expansion for the terms in brackets, we get:

B_z = \frac{\mu_0}{4} I R^2 \left[ \frac{1}{R^3} \left( 1 - \frac{3}{2} \frac{(z-m)^2}{R^2} + \frac{15}{8} \frac{(z-m)^4}{R^4} + 1 - \frac{3}{2} \frac{(z+m)^2}{R^2} + \frac{15}{8} \frac{(z+m)^4